hdu 4612 Warm up 桥缩点

4612 Warm hdu up

题目：给出一个图，添加一条边之后，问能够在新图中得到的最少的桥的数量。

分析：我们可以双联通分量进行缩点，原图变成了一棵树。问题变成了：求树中添加一条边之后，使得不在圈的边最少。显然求一边直径，用总边数减掉最长路上的边数就是答案。注意数据存在重边的情况。

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

#define debug puts("here")
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define foreach(i,vec) for(unsigned i=0;i<vec.size();i++)
#define pb push_back
#define RD(n) scanf("%d",&n)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w)
#define All(vec) vec.begin(),vec.end()
#define MP make_pair
#define PII pair<int,int>
#define PQ priority_queue
#define cmax(x,y) x = max(x,y)
#define cmin(x,y) x = min(x,y)
#define Clear(x) memset(x,0,sizeof(x))
/*

#pragma comment(linker, "/STACK:1024000000,1024000000")

int size = 256 << 20; // 256MB
char *p = (char*)malloc(size) + size;
__asm__("movl %0, %%esp
" :: "r"(p) );

*/

char IN;
bool NEG;
inline void Int(int &x){
    NEG = 0;
    while(!isdigit(IN=getchar()))
        if(IN=='-')NEG = 1;
    x = IN-'0';
    while(isdigit(IN=getchar()))
        x = x*10+IN-'0';
    if(NEG)x = -x;
}
inline void LL(ll &x){
    NEG = 0;
    while(!isdigit(IN=getchar()))
        if(IN=='-')NEG = 1;
    x = IN-'0';
    while(isdigit(IN=getchar()))
        x = x*10+IN-'0';
    if(NEG)x = -x;
}

/******** program ********************/

const int MAXN = 200005;
const int MAXM = 1000005;

vector<int> adj[MAXN];
int po[MAXN],tol;
int low[MAXN],dfn[MAXN],tim;
bool use[MAXN];
int fa[MAXN];
int id[MAXN],ha[MAXN];

struct Edge{
    int y,id,next;
}edge[MAXM<<1];

inline void add(int x,int y,int i){
    edge[++tol].y = y;
    edge[tol].id = i;
    edge[tol].next = po[x];
    po[x] = tol;
}

int findSet(int x){
    if(x!=fa[x])
        fa[x] = findSet(fa[x]);
    return fa[x];
}

void dfs(int x,int fid){
    low[x] = dfn[x] = ++ tim;
    for(int i=po[x];i;i=edge[i].next){
        int y = edge[i].y;
        int id = edge[i].id;
        if(id==fid)continue;
        if(!dfn[y]){
            dfs(y,id);
            cmin( low[x],low[y] );
            if(low[y]<=dfn[x]){
                int px = findSet(x);
                int py = findSet(y);
                if(px!=py)fa[px] = py;
            }
        }else
            cmin( low[x],dfn[y] );
    }
}

int MAX,root;

void dfsR(int x,int sz){
    use[x] = true;
    if(sz>MAX){
        MAX = sz;
        root = x;
    }
    foreach(i,adj[x])
        if(!use[adj[x][i]])
            dfsR(adj[x][i],sz+1);
}

int main(){

#ifndef ONLINE_JUDGE
    freopen("1002.in","r",stdin);
    //freopen("sum.out","w",stdout);
#endif

    int size = 256 << 20; // 256MB
    char *p = (char*)malloc(size) + size;
    __asm__("movl %0, %%esp
" :: "r"(p) );

    int x,y,n,m;
    while(1){
        Int(n);Int(m);
        if(!n&&!m)break;

        Clear(po);
        tol = 0;
        rep1(i,m){
            Int(x);Int(y);
            add(x,y,i);
            add(y,x,i);
        }

        rep1(i,n){
            adj[i].clear();
            fa[i] = i;
        }
        Clear(dfn);
        tim = 0;
        rep1(i,n)
            if(!dfn[i])
                dfs(i,0);

        Clear(id);
        int tot = 0;
        rep1(i,n){
            int px = findSet(i);
            if(!id[px])id[px] = ++tot;
            ha[i] = id[px];
        }
        rep1(x,n){
            int px = ha[x];
            for(int i=po[x];i;i=edge[i].next){
                int py = ha[ edge[i].y ];
                if(px==py)continue;
                adj[px].pb(py);
                adj[py].pb(px);
            }
        }

        Clear(use);
        MAX = 0;
        dfsR(1,1);

        Clear(use);
        MAX = 0;
        dfsR(root,1);

        printf("%d
",tot-MAX);
    }

    return 0;
}